# Principle of Powers problem

• May 28th 2008, 11:51 AM
lepraconn
Principle of Powers problem
Hi, first post:
I'm helping a student work a radical equation and have a question. Here is the problem:

√2X-5 = 1 + √X - 3 (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)

Step 2: (√2x – 5)2 = (1 + √x-3)2

Step 3: 2x – 5 = 1 = 2√x-3 + (x -3 ) à My question is HOW did they get the 2 in front of the radical sign!!!

Step 4: x – 3 = 2√x-3 à so I’m really lost on this one.

The answer is x = 7 or x = 3 but WHAT did they do between step 2 and step 3 to get the 2 in front of the radical sign?

Thanks.

• May 28th 2008, 11:59 AM
Moo
Hello,

Quote:

Originally Posted by lepraconn
Hi, first post:
I'm helping a student work a radical equation and have a question. Here is the problem:

√2X-5 = 1 + √X - 3 (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)

Step 2: (√2x – 5)2 = (1 + √x-3)2

Step 3: 2x – 5 = 1 = 2√x-3 + (x -3 ) à My question is HOW did they get the 2 in front of the radical sign!!!

~~

$\left(1+\sqrt{x-3}\right)^2=(a+b)^2=a^2+{\color{red}2}ab+b^2$

:)
• May 29th 2008, 01:06 AM
earboth
Quote:

Originally Posted by lepraconn
Hi, first post:
I'm helping a student work a radical equation and have a question. Here is the problem:

√(2X-5) = 1 + √(X - 3)
(2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)
...

I'll do this question step by step so you have an example how to do the next 166 questions of the same type:

1. In general: To get rid of the square-root you have to square - but correctly:

(a * b)² = a² * b²

(a+b)² = a² + 2ab + b²

(a-b)² = a² - 2ab + b²

2. The tactics in solving such an equation is: Square - isolate remainings roots - square - isolate .... until there isn't any root left:

$\sqrt{2x-5} = 1+\sqrt{x-3}~,~x \geq 3$ ........ square both sides

$2x-5=1+2 \cdot 1 \cdot \sqrt{x-3} + x-3$ ........ collect like terms and isolate the root on the RHS of the equation

$x-3 = 2\sqrt{x-3}$ ........ square both sides
$x^2-6x+9=4(x-3)$ ........ collect like terms

$x^2-10x+21=0$........ use the formula to solve this quadratic equation. I've got:

$x = 3 ~\vee~ x= 7$

Since squaring is not an equivalent transformation you have to plug in the results into the original equation and check if the solutions satisfy the equation:

1. $x = 3~\implies~ \sqrt{2 \cdot 3-5} = 1+\sqrt{3-3} ~\implies~ \sqrt{1} = 1$

2. $x = 7 ~\implies~ \sqrt{2 \cdot 7-5} = 1+\sqrt{7-3} ~\implies~ \sqrt{9} = 1 + \sqrt{4}$

Obviously both results are solutions of the equation.
• May 29th 2008, 01:12 AM
earboth
I've demonstrated the more usual way to solve such a type of questions. With your question you may use a short cut at this point:

$x-3 = 2\sqrt{x-3}$

Use the substitution: $\sqrt{x-3} =y$

Then the equation becomes:

$y^2 = 2y~\implies~y(y-2)=0~\implies~y=0~\vee~y=2$

That means:

$\sqrt{x-3}=0~\implies~x=3$

$\sqrt{x-3}=2~\implies~x-3=4~\implies~x=7$
• May 29th 2008, 04:09 AM
Soroban
Hello, lepraconn!

Welcome abarod!

Quote:

$\sqrt{2x-5} \:= \:1 + \sqrt{x-3}$

Step 2: . $\left(\sqrt{2x - 5}\right)^2 \:= \:\left(1 + \sqrt{x-3}\right)^2$

Step 3: . $2x - 5 \:= \:1 + 2\sqrt{x-3} + (x-3)$

My question is HOW did they get the 2 in front of the radical sign?

I don't suppose you even tried to square it out, did you?

$\left(1 + \sqrt{x-3}\right)\left(1 + \sqrt{x-3}\right) \;=\;1^2 + \sqrt{x-3} + \sqrt{x-3} + \left(\sqrt{x-3}\right)^2$

. . Got it?